3 Answers
3

For at least one choice of a constant k > 0, you can find a constant a such that the inequality f(x) < k g(x) holds for all x > a.

Note that O(g) is the set of all functions for which this condition holds.

f ∈ o(g) says, essentially

For every choice of a constant k > 0, you can find a constant a such that the inequality f(x) < k g(x) holds for all x > a.

Once again, note that o(g) is a set.

In Big-O, it is only necessary that you find a particular multiplier k for which the inequality holds beyond some minimum x.

In Little-o, it must be that there is a minimum x after which the inequality holds no matter how small you make k, as long as it is not negative or zero.

These both describe upper bounds, although somewhat counter-intuitively, Little-o is the stronger statement. There is a much larger gap between the growth rates of f and g if f ∈ o(g) than if f ∈ O(g).

One illustration of the disparity is this: f ∈ O(f) is true, but f ∈ o(f) is false. Therefore, Big-O can be read as "f ∈ O(g) means that f's asymptotic growth is no faster than g's", whereas "f ∈ o(g) means that f's asymptotic growth is strictly slower than g's". It's like <= versus <.

More specifically, if the value of g(x) is a constant multiple of the value of f(x), then f ∈ O(g) is true. This is why you can drop constants when working with big-O notation.

However, for f ∈ o(g) to be true, then g must include a higher power of x in its formula, and so the relative separation between f(x) and g(x) must actually get larger as x gets larger.

To use purely math examples (rather than referring to algorithms):

The following are true for Big-O, but would not be true if you used little-o:

x^2 ∈ O(x^2)

x^2 ∈ O(x^2 + x)

x^2 ∈ O(200 * x^2)

The following are true for little-o:

x^2 ∈ o(x^3)

x^2 ∈ o(x!)

ln(x) ∈ o(x)

Note that if f ∈ o(g), this implies f ∈ O(g). e.g. x^2 ∈ o(x^3) so it is also true that x^2 ∈ O(x^3), (again, think of O as <= and o as <)

Yes-- the difference is in whether the two functions may be asymptotically the same. Intuitively I like to think of big-O meaning "grows no faster than" (i.e. grows at the same rate or slower) and little-o meaning "grows strictly slower than".
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PhilSep 1 '09 at 20:38

@Phil Good wording. I worked that into my answer.
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Tyler McHenrySep 1 '09 at 20:46

2

One thing I would change about your answer is f=O(g). This is incorrect. It should be f \in O(g). If you use equality, I can do silly things like prove 1=2 since 1n and 2n both equal O(n).
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jlund3Oct 24 '12 at 1:17

Big-O is to little-o as ≤ is to <. Big-O is an inclusive upper bound, while little-o is a strict upper bound.

For example, the function f(n) = 3n is:

in O(n^2), o(n^2), and O(n)

not in O(lg n), o(lg n), or o(n)

Analogously, the number 1 is:

≤ 2, < 2, and ≤ 1

not ≤ 0, < 0, or < 1

Here's a table, showing the general idea:

(Note: the table is a good guide but its limit definition should be in terms of the superior limit instead of the normal limit. For example, 3 + (n mod 2) oscillates between 3 and 4 forever. It's in O(1) despite not having a normal limit, because it still has a lim sup: 4.)

I recommend memorizing how the Big-O notation converts to asymptotic comparisons. The comparisons are easier to remember, but less flexible because you can't say things like n^O(1) = P.

I find that when I can't conceptually grasp something, thinking about why one would use X is helpful to understand X. (Not to say you haven't tried that, I'm just setting the stage.)

[stuff you know]A common way to classify algorithms is by runtime, and by citing the big-Oh complexity of an algorithm, you can get a pretty good estimation of which one is "better" -- whichever has the "smallest" function in the O! Even in the real world, O(N) is "better" than O(N^2), barring silly things like super-massive constants and the like.[/stuff you know]

Let's say there's some algorithm that runs in O(N). Pretty good, huh? But let's say you (you brilliant person, you) come up with an algorithm that runs in O(N/loglogloglogN). YAY! Its faster! But you'd feel silly writing that over and over again when you're writing your thesis. So you write it once, and you can say "In this paper, I have proven that algorithm X, previously computable in time O(N), is in fact computable in o(n)."

Thus, everyone knows that your algorithm is faster --- by how much is unclear, but they know its faster. Theoretically. :)